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I have heard whisperings that if we have a turing machine that is allowed to compute infinitely many steps in finite time, then we can solve the halting problem.

This made me wonder, if we have such a TM, do Godel's incompleteness theorems then not hold anymore?

My intuition is: If the TM can make an infinite amount of computations, it can simply check the truth-value of any formula, by checking all natural numbers. e.g. if we have a formula $\forall x \phi (x)$, the infinite-TM can simply check for every natural number $n$ whether $\phi(n)$ holds.

user56834
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1 Answers1

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There is some ambiguity in how to define these new machines, essentially based on the question whether we can nest the "bursts of doing infinitely many steps in some finite time" or not. In the weakest interpretation, we essentially just get an oracle Turing machine with access to the Halting problem. In the strongest, we get the Infinite Time Turing Machines (ITTM).

Both machines can easily decide the ordinary Halting problem. However, the proof of non-computability of the Halting problem uses very little prerequisites. In particular, it shows that none of our machines can solve their own Halting problems.

So while an ITTM can decide truth of all basic arithmetic formulas (by just searching through the natural numbers whenever required), they cannot decide all $\Sigma^1_2$-formulas. And to properly define and work with ITTMs, we need $\Sigma^1_2$-formulas. So we have just pushed undecidability further out, but we have not removed it.

A $\Sigma^1_2$-formula is one of the form $$\exists A \subseteq \mathbb{N} \ \ \forall B \subseteq \mathbb{N} \ \ \phi(A,B)$$ where $\phi$ is a formula which contains only quantification over numbers, but not over sets of numbers.

Arno
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